I try to pass to my view sample Blade the result of a method that is in:
App / Conversations / bot.php
private function searchCompatibility()
{
$users = Products::all();
return View::make('sample')->with('products', $products);
}
In my routes I want to send the result of my method to my view, how can I send it to call, what is the correct syntax? Or can it only be from a controller?
You have declared $users and you pass $products in your return statement
Change the variable $users to $products
You wrongly define your variable. you define $users but you pass as $product
private function searchCompatibility()
{
$products = Products::all();
return view('sample', compact('products'));
}
Related
I want to get all user's online friends, how can I call a custom model function inside the eloquent condition?
this is my code
$friends = $user->friends()->where(function (Builder $query){
$query->where('friend', 'yes');
})
->get();
and this is my function in model
public function getIsOnlineAttribute(): bool
{
// check if the user is online or not
return $this->is_online;
}
I can access is_online after eloquent by foreach, but in my case, I want to check everything in one step ( inside where condition in eloquent). how can I do that???
You can't use conditions for eloquent accessors, in this case you can use (assume 1 is database column value):
$friends = $user->friends()->where('is_online', 1)->get();
or
$friends = $user->friends()->whereIsOnline(1)->get();
or you can create eloquent scope on your model:
public function scopeIsOnline($query) {
$query->where('is_online',1);
}
and you can use this eloquent scope on your controller in this way:
$friends = $user->friends()->isOnline()->get();
this worked for me :)
$friends = $user->friends()
->simplePaginate()
->reject(function ($friend) {
return $friend->is_online === false;
});
I'm trying to get all categories of restaurants.
My models:
Restaurant
public function categories()
{
return $this->belongsToMany(Category::class,'restaurant_categories_relation','restaurant_id');
}
Category
public function restaurants()
{
return $this->belongsToMany(Restaurant::class,'restaurant_categories_relation', 'category_id');
}
In my controller:
$restaurants = Restaurant::where('district_id', $request->district)->paginate(8);
$categories = $restaurants-> ????;
Please help me do this, thanks!
You could use has() like :
Category::has('restaurants')->get();
That will return the categories who are related with the restaurants.
Try also the use of whereHas like :
$users = Category::whereHas('restaurants', function($q){
$q->->where('district_id', $request->district)->paginate(8);
})->get();
Since you've already a Collection we can't query the categories so I suggest adding a function to scope that inside the Restaurant model like :
public static function getCategoriesOfRestaurants($restaurants)
$categories = [];
foreach($restaurants as $restaurant){
array_push( $categories, $restaurant->categories->pluck('id')->toArray());
}
return Category::WhereIn('id', array_unique($categories))->get();
}
Then just call it when you get the $restaurants collection :
$restaurants = Restaurant::with("categories")->where('district_id', $request->district)->paginate(8);
$categories = Restaurant::getCategoriesOfRestaurants($restaurants);
Note: The use of with("categories") when getting the collection will query All the related categories in the first query so the foreach loop will not generate any extra query just looping through the already fetched data, and finally we will get the collection of categories in the return statement.
use with() method for eagerloading, that provides you get all categories in a single query
$restaurants = Restaurant::with("categories")->where('district_id', $request->district)->paginate(8);
foreach($restaurants as $restaurant){
foreach($restaurant->categories as $category)
{{$category}}
}
}
if you want to use categories outside of the loop, then assign these categories to a variable
foreach($restaurants as $restaurant){
$categories = $restaurant->categories;
}
// do something with $categories
Your relation should be
Restruant.php
public function categories() {
return $this->belongsToMany(Category::class,'restaurant_categories_relation','restaurant_id','category_id');
}
then in you controller method just wirte
$restruants = Restruant::with('categories')->get();
It should return you collection of all restruants with all related categories.
I want to send variable $typeid to function categories to use it in query is there a way Knowing that when I try to use new instance of class in my controller like that:
$cat= new Main_category();
$categories = $cat->categories()->get();
it returns empty array
the following code is working well when I manually add the typeid inside the model function I want to have it as a variable sent from controller
controller:
$categories = Main_category::with('categories')->get();
Model
public function categories()//($typeid)
{
$query = $this->hasMany(Category::class, 'main_cat_id')
->join('category_type','category_type.cat_id','=', 'categories.cat_id')
->join('main_categories','main_categories.main_cat_id','=', 'categories.main_cat_id')
->where('category_type.type_id', '1'); // I want to use $typeid here
return $query;
}
I am not sure whether you can pass your variable in eloquent relationship methods by using with method or not. But you can add a where clause in controller.
Main_category::with(['categories' => function($query) use($typeid) {
$query->where('category_type.type_id', $typeid);
}])->get();
Or you can create a query scope for model too.
in Model
public function scopeWithCategories($query, $typeid) {
return $query->with(['categories' => function($query) use($typeid) {
$query->where('category_type.type_id', $typeid);
}]);
}
and finally in Controller
Main_category::withCategories($typeid)->get();
In my Post Model
public function user()
{
return $this->belongsTo('App\User');
}
And in the User Model
public function posts()
{
return $this->hasMany('App\Post');
}
Now I am trying to get the comments of a specific user
$user= User::where('name', 'like', '%Mat%')->first();
return $user->posts->comment;
But it shows
Property [comment] does not exist on this collection instance.
The user has many posts which therefore returns a collection, you will need to loop over this to get your comments out. I.e.
$user = User::where('name', 'like', '%Mat%')->first();
$user->posts->each(function($post) {
echo $post->comment;
});
See the documentation on Laravel Collections
I think you can try this :
$user= User::with('post')->where('name', 'like', '%Mat%')->get();
$postComment = array();
foreach($user->post as $post){
$postComment = $post->comment;
}
return $postComment;
Hope this help for you !!!
If you want to have all comments you can use the following code:
$comments = [];
$user = User::where('name', 'like', '%Mat%')->with(['post.comment' => function($query) use (&$comments) {
$comments = $query->get();
}])->first();
return $comments;
Property [comment] does not exist on this collection instance.
The above error occurs because the Posts function returns a collection. Now you will have to traverse through each element of the collection.
Since, you are returning $user->posts()->comment, I am assuming you need it in the form of an array and don't have to simply echo them out, one by one. So you can store them all in an array & then process it whatever whay you like.
$comments = array();
$user->posts()->each(function $post){
$comments = $post->comment;
}
return $comments;
For greater insight, into this collection function read:
https://laravel.com/docs/5.4/collections#method-each
Need help on this simple code, what code should I replace at
$users = User::all(); so I can conditional chaining the scope method and paginate it at the end?
I tried initiate the User object with $users = new User(); and there is error, Trying to get property of non-object error when using at VIEW.
public function index()
{
// user search
$name = $this->request->name;
$email = $this->request->email;
$users = User::all();
if (!empty($name)) {
$users->name($name);
}
if (!empty($email)) {
$users->email($email);
}
$users->paginate(5);
return view('admin.users.index',compact('users'));
}
Thanks in advance
$users = DB::table('users');
$users = empty($email) ? $users->paginate(5) : $users->whereEmail($email)->paginate(5);
You do not need to check for a name if all users have an email, because email is always unique. That way to do the task is faster and more convinient.
In case, if you need to check name or anything else, you can use this:
$users = DB::table('users');
$users = empty($email) ? $users : $users->whereEmail($email);
$users = empty($name) ? $users : $users->whereName($name);
$users = $users->paginate(5);
Try this:
$users = new User;
if( !empty($name) )
$users->whereName($name);
if( !empty($email) )
$users->whereEmail($email);
$users =$users->paginate(5);
I found the answer from the following link
https://stackoverflow.com/a/21739314/417899
Eg code:
$users = new User;
if (!empty($name)) {
$users = $users->name($name);
}
$users = $users ->paginate(5);